Question

If the Laplace transform of a function \( f(t) \) is given by \[ \frac{2s + 1}{(s + 1)(s + 2)} \] then \( f(0) \) is equal to ______

• A. 2
• B. 4
• C. -4
• D. \( 3e - 1 \)

Hint:

Use the initial value theorem \( f(0) = \lim_{s \to \infty} sF(s) \) when given the Laplace transform of \( f(t) \).

Answer 1

The Correct Option is A

We are given the Laplace transform: \[ F(s) = \frac{2s + 1}{(s + 1)(s + 2)} \]
To find \( f(0) \), recall the initial value theorem: \[ f(0) = \lim_{s \to \infty} sF(s) \]
Substitute: \[ f(0) = \lim_{s \to \infty} s \cdot \frac{2s + 1}{(s + 1)(s + 2)} \]
Multiply numerator and simplify: \[ f(0) = \lim_{s \to \infty} \frac{s(2s + 1)}{s^2 + 3s + 2} \]
The leading term in the numerator is \( 2s^2 \), and in the denominator is also \( s^2 \):
\[ f(0) = \lim_{s \to \infty} \frac{2s^2 + s}{s^2 + 3s + 2} \rightarrow \frac{2}{1} = 2 \]